Integration of dirac function explanation

Question

I have a problem that need your help. I have a gray image. We denotes $I(x)$ is gray level of a pixel in the image and $f(z)$ is a function of $z$(ie: histogram function...)-where $z$ is the set of gray level values in the set {1, 2, ..., 256} in $Z$, $\delta(.)$ is as smooth Gaussian kernel. The $V$ function is express as below figure. Hence I have a two questions that need you explain:

1. The $V$ function can express as convolution $V=(\delta*f)(I(x))$, Is it right?
2. Please explain the $V$ formula. What does this mean? As my knowledge, assume that $f(z)$ is histogram of gray level 1 to 256. and $\delta(z-I(x)$ is dirac function at point $x$, hence the multiply of $\delta(z-I(x)$ and $f(z)$ will return a narrow distribution around the pixel $x$. The integrate of them that means we will do it for every gray level in $Z$ domain. Hence, $V$ return a narrow distribution of set of gray level around the pixel $x$. Is it correct?

Paper reference: http://jgmalcolm.com/pubs/others/rathi_seeing_unseen.pdf

Update: The implementation: Assume $f(z)$ can denotes f_z=[0.01 0.02 ..0.001] with size 256x1, Image I is 3x3 the V function can implement as

V=f_z(I+1) %where I is image and f_z size is 256x1 - hence V size is 3x3

Please check my implementation is correct or not?

I=[1 1 2; 256 1 2; 256 256 256];

So the result is

V=[0.01 0.01 0.02; 0.001 0.01 0.02; 0.001 0.001 0.001]

Answer 1

The $V$ function can be expressed as convolution $V=(\delta*f)(I(x))$, is that right?

Yes, that's correct. Furthermore, since $\delta * f = f$ for any function $f$, this integral gives us exactly $$ V = \int_{\mathcal Z} \delta(z - I(x)) \cdot f(z)\,dz = (\delta*f)(I(x)) = f(I(x)) $$ So, it is not just "a narrow distribution of set of gray level around the pixel $x$". It is, in fact, exactly the value of the gray level at that pixel.

This, however, assumes that $\delta$ is precisely the Dirac-delta distribution. Using approximations (such as the $\delta_\epsilon$ in the paper) gives us a weighted average of the values of $f$ nearby that approaches $f(I(x))$ as $\epsilon \to 0$.

Answer 2

$I(x)$ is the gray value at the pixel $x$; $f(z)$ is some unspecified function of the gray value (it could be a histogram frequency, but that's unsure). $V$ is a function of $x$ taking values in the image of $f$.

Has $\delta$ denoted a Dirac delta, you would have had $$V(x)=f(I(x)),$$ i.e. take the gray level at $x$, let $z=I(x)$, and compute $f(z)$.

In the case of a (narrow) Gaussian, instead of $f(z)$, you will compute an average of $f$ in the vicinity of $z=I(x)$, i.e. a moothed version of $f$.

If $f$ is indeed the histogram of image $I$, the final $V$ output will show the frequency of the pixels in $I$, i.e. bright values for the most common gray scales (usually the background).